Question: ${\sqrt[3]{32} = \text{?}}$
$\sqrt[3]{32}$ is the number that, when multiplied by itself three times, equals $32$ First break down $32$ into its prime factorization and look for factors that appear three times. So the prime factorization of $32$ is $2\times 2\times 2\times 2\times 2$ Notice that we can rearrange the factors like so: $32 = 2 \times 2 \times 2 \times 2 \times 2 = (2\times 2\times 2) \times 2\times 2$ So $\sqrt[3]{32} = \sqrt[3]{2\times 2\times 2} \times \sqrt[3]{2\times 2}$ $\sqrt[3]{32} = 2 \times \sqrt[3]{2\times 2}$ $\sqrt[3]{32} = 2 \sqrt[3]{4}$